Derivatives of Rational Functions

Remember that a rational function $h(x)$ can be expressed in such a way that $h(x)=\frac{f(x)}{g(x)},$ where $f(x)$ and $g(x)$ are polynomial functions. Using this basic fundamental, we can find the derivatives of rational functions. Let's check how to do it.

The derivative of a rational function may be found using the quotient rule:

Let ${h(x)=\frac{f(x)}{g(x)}},$ then ${h'(x)=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}}.$

We start with the basic definition of a derivative that is

$$\displaystyle \dfrac { dh(x) }{ dx } =\lim _{ \Delta x\rightarrow 0 }{ \frac { h(x+\Delta x)-h(x) }{ \Delta x } }.$$

Since we have $\frac{f(x)}{g(x)} = h(x)$, we can rewrite it as

$$\displaystyle \dfrac { dh(x) }{ dx } =\lim _{ \Delta x\rightarrow 0 }{ \dfrac { \frac { f(x+\Delta x) }{ g(x+\Delta x) } -\frac { f(x) }{ g(x) } }{ \Delta x } }.$$

Simplifying this, we have

$$\begin{aligned} \frac { dh(x) }{ dx } &=\lim _{ \Delta x\rightarrow 0 }{ \frac { f(x+\Delta x)g(x)-f(x)g(x+\Delta x) }{ \Delta xg(x)g(x+\Delta x) } } \\\\ &= \lim _{ \Delta x\rightarrow 0 }{ \frac { 1 }{ g(x)g(x+\Delta x) } } \lim _{ \Delta x\rightarrow 0 }{ \frac { f(x+\Delta x)g(x)-f(x)g(x+\Delta x) }{ \Delta x } } \\\\ &=\frac { 1 }{\big(g(x)\big)^2} \lim _{ \Delta x\rightarrow 0 }{ \frac { f(x+\Delta x)g(x)-f(x)g(x+\Delta x) }{ \Delta x } }. \end{aligned}$$

Adding and subtracting $f(x)g(x)$ in the numerator, we have

$$\displaystyle \frac { dh(x) }{ dx } =\frac { 1 }{ { \big(g(x)\big) }^{ 2 } } \lim _{ \Delta x\rightarrow 0 }{ \frac { f(x+\Delta x)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+\Delta x) }{ \Delta x } }$$

Manipulating a little, we have

$$\begin{aligned} \frac { dh(x) }{ dx } &=\frac{1}{{ g(x) }^{2}} \lim _{ \Delta x\rightarrow 0 }{ \left(g(x)\bigg(\frac { f(x+\Delta x)-f(x) }{ \Delta x } \bigg)-f(x)\bigg(\frac { g(x+\Delta x)-g(x) }{ \Delta x } \bigg)\right) } \\\\ &=\dfrac { \displaystyle \left(g(x)\bigg(\lim _{ \Delta x\rightarrow 0 }{ \frac { f(x+\Delta x)-f(x) }{ \Delta x } } \bigg)-f(x)\bigg(\lim _{ \Delta x\rightarrow 0 }{ \frac { g(x+\Delta x)-g(x) }{ \Delta x } } \bigg)\right) }{ { g(x) }^{ 2 } }. \end{aligned}$$

Finally, we get

$$\displaystyle \boxed { \dfrac { dh(x) }{ dx } =\dfrac { f'(x)g(x)-g'(x)f(x) }{ { \big(g(x)\big) }^{ 2 } } }.\ _\square$$

Find $\frac{d}{dx}\left(\frac{3x^3-x-2}{2x}\right).$

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Using the formula above, let $3x^3-x-2=f(x)$ and $2x=g(x).$ Then

$$\begin{aligned} \frac{d}{dx} h(x) &=\frac{(2x)(9x^2-1)-(3x^3-x-2)(2)}{(2x)^2}\\ &=\frac{18x^3-2x-6x^3+2x+4}{4x^2}\\ &=\frac{3x^3+1}{x^2}.\ _\square \end{aligned}$$

Doing differentiation for a rational term is quite complicated and confusing when the expressions are very much complicated. In such cases, you can assume the numerator as one expression and the denominator as one expression and find their separate derivatives. Now write the combined derivative of the fraction using the above formula and substitute directly so that there will be no confusion and the chances of doing mistakes will be reduced. The following few examples illustrate how to do this:

If $y = \frac{a - x}{a + x}\ (x \neq -a),$ then find $\frac{dy}{dx}$.

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Write $u(x) = a - x \implies u'(x) = -1$ and $v(x) = a + x \implies v'(x) = 1$ so that $y = \frac{u(x)}{v(x)}:$

$$\begin{aligned} \dfrac{dy}{dx} & = \dfrac{v(x)u'(x) - v'(x)u(x)}{\big(v(x)\big)^2} \\\\ & = \dfrac{(a + x)(-1) - (a - x)(1)}{(a + x)^2} \\\\ & = \dfrac{-2a}{a^2 + 2ax + x^2} .\ _\square \end{aligned}$$

If $y = \frac{px^2 + qx + r}{ax + b}\ \big(|a| + |b| \neq 0\big),$ then find $\frac{dy}{dx}$.

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Write $u(x) = px^2 + qx + r \implies u'(x) = 2px + q$ and $v(x) = ax + b \implies v'(x) = a$ so that $y = \frac{u(x)}{v(x)}:$

$$\begin{aligned} \dfrac{dy}{dx} & = \dfrac{d}{dx} \left(\dfrac{u(x)}{v(x)}\right)\\\\ & = \dfrac{v(x)u'(x) - v'(x)u(x)}{\big(v(x)\big)^2} \\\\ & = \dfrac{(ax + b)(2px + q) - (a)(px^2 + qx + r)}{(ax + b)^2} \\\\ & = \dfrac{apx^2 + 2bpx + bq - ar}{a^2x^2 + 2abx + b^2}.\ _\square \end{aligned}$$

If $y = \frac{1}{ax^2 + bx + c}\ \big(|a| + |b| + |c| \neq 0\big),$ then find $\frac{dy}{dx}$.

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Write $u(x) = 1 \implies u'(x) = 0$ and $v(x) = ax^2 + bx + c \implies v'(x) = 2ax + b$ so that $y = \frac{u(x)}{v(x)}:$

$$\begin{aligned} \dfrac{dy}{dx} & = \dfrac{d}{dx} \left(\dfrac{u(x)}{v(x)}\right)\\\\ & = \dfrac{v(x)u'(x) - v'(x)u(x)}{\big(v(x)\big)^2} \\\\ & = \dfrac{(ax^2 + bx + c)(0) - (2ax + b)(1)}{(ax^2 + bx + c)^2} \\\\ & = \dfrac{-(2ax + b)}{(ax^2 + bx + c)^2} .\ _\square \end{aligned}$$

If $y = \frac{ax + b}{cx + d}\ \big(|c| + |d| \neq 0\big),$ then find $\frac{dy}{dx}$.

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Write $u(x) = ax + b \implies u'(x) = a$ and $v(x) = cx + d \implies v'(x) = c$ so that $y = \frac{u(x)}{v(x)}:$

$$\begin{aligned} \dfrac{dy}{dx} & = \dfrac{d}{dx} \left(\dfrac{u(x)}{v(x)}\right)\\\\ & = \dfrac{v(x)u'(x) - v'(x)u(x)}{\big(v(x)\big)^2} \\\\ & = \dfrac{(cx + d)(a) - (c)(ax + b)}{(cx + d)^2} \\\\ & = \dfrac{ad - bc}{(cx + d)^2} .\ _\square \end{aligned}$$

If $y = \frac{1 - x\sqrt{x}}{1 + x\sqrt{x}}\ (x > 0),$ then find $\frac{dy}{dx}$.

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Write $u(x) = 1 - x\sqrt{x} \implies u'(x) = 0 - \sqrt{x} - \frac{x}{2\sqrt{x}} = - \frac{3\sqrt{x}}{2}$ and $v(x) = 1 + x\sqrt{x} \implies v'(x) = 0 + \sqrt{x} + \frac{x}{2\sqrt{x}} = \frac{3\sqrt{x}}{2}.$

Now,

$$\begin{aligned} y & = \dfrac{u(x)}{v(x)} \\\\ \Rightarrow \dfrac{dy}{dx} & = \dfrac{v(x)u'(x) - v'(x)u(x)}{\big(v(x)\big)^2} \\\\ & = \dfrac{(1 + x\sqrt{x})\left(-\dfrac{3\sqrt{x}}{2}\right) - \left(\dfrac{3\sqrt{x}}{2}\right)(1 - x\sqrt{x})}{(1 + x\sqrt{x})^2} \\\\ & = \dfrac{-3\sqrt{x} - 3x^2 - (3\sqrt{x} - 3x^2)}{2(1 + x\sqrt{x})^2} \\\\ & = \dfrac{-6\sqrt{x}}{2(1 + x\sqrt{x})^2} \\\\ & = -\dfrac{3\sqrt{x}}{(1 + x\sqrt{x})^2} .\ _\square \end{aligned}$$